JIPMAT 2021 Question Paper

$$\left(\frac{1}{\sqrt{9} - \sqrt{8}} -\frac{1}{\sqrt{8} - \sqrt{7}}+\frac{1}{\sqrt{7} - \sqrt{6}}-\frac{1}{\sqrt{6} - \sqrt{5}}+\frac{1}{\sqrt{5} - \sqrt{4}} \right)$$ 

= ($$\sqrt{9\ }+\ \sqrt{\ 8}$$) - ($$\sqrt{8\ }+\ \sqrt{\ 7}$$) + ($$\sqrt{7\ }+\ \sqrt{\ 6}$$) - ($$\sqrt{6\ }+\ \sqrt{\ 5}$$) + ($$\sqrt{5\ }+\ \sqrt{\ 4}$$) 

= ($$\sqrt{9\ }+\ \sqrt{\ 4}$$)
= 3 + 2 
= 5

Let the total or max. marks be "T" and let the passing marks be "P" .
It is given : 25 % of T = P - 30
$$\ \dfrac{\ T}{4}=P-30$$

It is also given : 50% of T = P + 20
$$\ \dfrac{\ T}{2}=P+20$$

Therefore, P = 80 & T = 200.

Hence, the pass marks percentage is 40% .

As it is given second sum is twice the first sum.
Let the first total sum be X and second total sum be 2X.

First Distribution :  4 : 3 : 2 : 1
So, Naresh = 4X/10 = 0.4(X)
      Vipin = 3X/10 = 0.3(X)
      Bhupesh = 2X/10 = 0.2(X)
      Yogesh = X/10 = 0.1(X)

Second Distribution :  5 : 6 : 7 : 8.
So, Naresh = 5(2X)/26 =  0.38 (X)
     Vipin = 6(2X)/26 = 0.46(X)
     Bhupesh = 7(2X)/26 = 0.53(X)
     Yogesh = 8(2X)/26 = 0.61(X)

Therefore total distributions for each : 
Naresh = X(0.4+0.38) = 0.78(X)
Vipin = X(0.46+0.3) = 0.76(X)
Bhupesh = X (0.53+0.2) = 0.73(X)
Yogesh = X(0.61+0.1) = 0.71(X)

Therefore, the highest distribution is of Naresh.

As given three points are collinear : Slope between any two points on the line should remain same.
         (t, 2t), (-2, 6) and (3,1)

So, Slope = $$\dfrac{\ 6-1}{-2-3}\ =\left\{-1\right\}$$

Slope = $$\dfrac{\ 2t-1}{t-3}$$ = -1

2t - 1 = 3 - t

Therefore, t = $$\dfrac{4}{3}$$

$$(25.732^{2}-15.732^{2})$$ 
= $$(25+0.732)^2-(15+0.732)^2$$
= $$25^2+50(0.732)+0.732^2-15^2-0.732^2-30(0.732)$$
= 400 + 20(0.732)
= 414.64

It is said that the Entire work (W) is finished in 40 days .
First 35 days : 100 men worked .
Next 5 days : (100+100) men worked.

Let the amount of work done by each men per day be "K"
So, 100(K)(35) + 200(K)(5) = W 
So, K = $$\dfrac{\ W}{4500}$$

If he had not engaged additional men, then only 100 people would have only worked with efficiency "K" till the entire work is finished. So lets time taken to finish the work in this case be "T" : 
                            100(K)(T) = W 
                             T  =$$\dfrac{\ W}{100\left(K\right)}$$
                                 = 4500/100 
                                 = 45 days.

Therefore, the work could have been finished with a delay of (45-40) days = 5 days.

Let the amount of work done by each men per each day be X.
Let the amount of work done by each women per each day be Y.
Let the total amount of work be W. 

Given 4 men & 6 women finish job in 8 days : (4X + 6Y)(8) = W
Given 3 men and 7 women finish job in 10 days : (3X + 7Y)(10) = W

Upon solving above two equations, we get : X = 11 Y
Substituting the same in (4X + 6Y)(8) = W
    (44Y + 6Y)(8) = W
      Y = W/400 

Therefore, if only 10 women work on this job : 10Y(t) = W
Hence, t = 40 days.                                                                 

Let the amount of grams shopkeeper cheats for 1000 gm be X gm .

So, he sells only (1000-X) gms  to customer at a selling price given to be equal to CP.
The original Cost Price which would have costed shopkeeper =$$\dfrac{\ CP}{1000}\left(1000-X\right)$$

Therefore , Profit % =$$\ \dfrac{CP-\frac{\ CP}{1000}\left(1000-X\right)\ }{\dfrac{\ CP}{1000}\left(1000-X\right)\ }$$ x 100%

                              = $$\dfrac{\ X}{1000-X}\times100$$

        $$\dfrac{\ X}{1000-X}\times100\%$$ = $$6\dfrac{18}{47}\%$$
             
         $$\dfrac{\ X}{1000-X}=\dfrac{\ 3}{47}$$

X = 60 gms .

Therefore (1000-60) = 940 gm

Total time taken in journey = 110 mins = $$\dfrac{\ 11}{6}$$ hrs

Total time = $$\dfrac{\ 20}{V\ +\ 2}+\dfrac{\ 20}{V\ -2}$$ = $$\dfrac{\ 11}{6}$$ hrs

$$\dfrac{\ 11}{120}=\dfrac{\ 2V}{V^2-4}$$

$$11(V^2)-44=240(V)$$
$$11(V^2)-240(V)-44=0$$

Therefore, V = 22 kmph.

Let the number be "XY" .
Given the sum of digits =  X + Y = 8.........Equation 1

This XY can be written as : XY = Y(1) + X(10) 

Upon adding 18 to XY we get :  {10(X) + Y + 18}

Now, it is given this particular number is "YX" i.e it should be equal to {X + 10(Y)}.
Hence, 10(X) + Y + 18 = X + 10(Y)
9X - 9Y = -18
X - Y = (-2)..........Equation 2

By solving both equations : X = 3 & Y = 5.

Therefore, the number is 35.